Cooperative and Noncooperative Extraction in a Common Pool with Habit Formation

Abstract

The present paper considers the exploitation of a common-property, nonrenewable resource, by individuals subject to habit formation. We formalize their behavior by means of a utility function, depending on the difference between the individuals’ current consumption and the consumption level which they aspire, the latter being a weighted average of past consumptions in the population. We derive and compare the benchmark cooperative solution and a noncooperative Markov-perfect Nash equilibrium of the differential game. We investigate how the intensity, persistence and initial level of habits shape the cooperative and noncooperative solutions. We prove that habit formation may either mitigate or worsen the tragedy of the commons.

Appendix

1.1 Proof of Condition 1

Suppose that $q_{i}\left( t\right) =\left( c_{i}\left( t\right) -\alpha y\left( t\right) \right) /\left( 1-\alpha \right) =0$, for all i and t. Substituting $c_{i}\left( t\right) =\alpha y\left( t\right)$, for all i and t, (1) and (2), respectively, write

$$\begin{aligned} \dot{x}\left( t\right) =-\alpha ny\left( t\right) \end{aligned}$$

and

$$\begin{aligned} \dot{y}\left( t\right) =-\left( 1-\alpha \right) \beta ny\left( t\right) \text{. } \end{aligned}$$

From this, we can obtain

$$\begin{aligned} x\left( t\right) =x_{0}-\frac{\alpha }{\left( 1-\alpha \right) \beta }y_{0}\left( 1-\mathrm{e}^{-\left( 1-\alpha \right) \beta nt}\right) \end{aligned}$$

and

$$\begin{aligned} y\left( t\right) =y_{0}\mathrm{e}^{-\left( 1-\alpha \right) \beta nt}\text{. } \end{aligned}$$

From $y\left( t\right) \ge 0$, we verify that $c_{i}\left( t\right) \ge 0$, for all t. It is clear that $x\left( t\right) \ge 0$, for all t, if and only if $x_{0}-\frac{\alpha }{\left( 1-\alpha \right) \beta }y_{0}\ge 0$. $\square$

1.2 Proof of Proposition 1

By concavity of $u\left( \cdot \right)$, the cooperative solution is symmetric. Hence, substituting $c_{i}\left( t\right) =c\left( t\right)$, for all i and t, the social problem is to choose $c\left( \cdot \right)$ to maximize

$$\begin{aligned} \begin{array}{l} \int _{0}^{\infty }nu\left( q\left( t\right) \right) \mathrm{e}^{-\delta t}\mathrm{d}t\text {,}\\ \text {where:}\\ \dot{x}\left( t\right) =-nc\left( t\right) ,\quad x\left( 0\right) =x_{0}\text {,}\\ \dot{y}\left( t\right) =\beta n\left( c\left( t\right) -y\left( t\right) \right) ,\quad y\left( 0\right) =y_{0}\text {,}\\ q\left( t\right) \ge 0,\quad x\left( t\right) \ge 0\text {,} \end{array} \end{aligned}$$

where we write $q\left( t\right) =\left( c\left( t\right) -\alpha y\left( t\right) \right) /\left( 1-\alpha \right)$ to simplify the notations.

Define the current-value Hamiltonian

$$\begin{aligned} H\left( x,y,\lambda ,\mu ,\eta ,c\right) =n\left[ u\left( q\right) -\lambda c+\beta \mu \left( c-y\right) +\eta q\right] , \end{aligned}$$

where $\lambda$ and $\mu$ are costate variables associated with the states x and y, and $\eta$ is a Lagrangian multiplier associated with the feasibility constraint.

If a feasible control path $c\left( \cdot \right)$, with the corresponding states trajectories $x\left( \cdot \right)$ and $y\left( \cdot \right)$, is optimal, there exists (continuous with piecewise-continuous derivatives) costate variables $\lambda \left( \cdot \right)$ and $\mu \left( \cdot \right)$, and a (continuous) multiplier $\eta \left( \cdot \right)$ such that

$$\begin{aligned}&\frac{u^{\prime }\left( q\left( t\right) \right) }{1-\alpha }=\lambda \left( t\right) -\beta \mu \left( t\right) -\frac{\eta \left( t\right) }{1-\alpha },\quad \end{aligned}$$

(5)

$$\begin{aligned}&\eta \left( t\right) \ge 0,\quad q\left( t\right) \ge 0\quad \text{ and }\,\eta \left( t\right) q\left( t\right) =0 \end{aligned}$$

(6)

$$\begin{aligned}&\dot{\lambda }\left( t\right) =\delta \lambda \left( t\right) , \end{aligned}$$

(7)

$$\begin{aligned}&\dot{\mu }\left( t\right) =\left( \delta +\beta n\right) \mu \left( t\right) +\frac{\alpha }{1-\alpha }n\left( u^{\prime }\left( q\left( t\right) \right) +\eta \left( t\right) \right) , \end{aligned}$$

(8)

$$\begin{aligned}&\lim _{t\rightarrow \infty }\mathrm{e}^{-\delta t}\left( \lambda \left( t\right) x\left( t\right) +\mu \left( t\right) y\left( t\right) \right) =0. \end{aligned}$$

(9)

Now, $f\left( z\right)$ being such that

$$\begin{aligned} \Theta \left( f\left( z\right) \right) \equiv \int _{0}^{f\left( z\right) }\sigma \left( s\right) \mathrm{d}s=\frac{\delta }{n}z,\quad \text{ for } \text{ all }\,z\ge 0, \end{aligned}$$

for all t, let $c\left( t\right)$ be such that

$$\begin{aligned} q\left( t\right) =\frac{c\left( t\right) -\alpha y\left( t\right) }{1-\alpha }=f\left( x\left( t\right) -\frac{\alpha }{\left( 1-\alpha \right) \beta }y\left( t\right) \right) , \end{aligned}$$

where $x\left( t\right)$ and $y\left( t\right)$ are the corresponding states trajectories.

(1) Feasibility We first show, in Lemmas 1 and 2, that the proposed control path $c\left( \cdot \right)$ is feasible.

Lemma 1

For all $z>0$, $f\left( z\right) >0$, and $f\left( 0\right) =0$.

Proof

If $z=0$, $f\left( 0\right) =0$ follows from $\Theta \left( 0\right) =0$. Likewise, $\lim _{z\rightarrow \infty }f\left( z\right) =\infty$ follows from $\lim _{q\rightarrow \infty }\Theta \left( q\right) =\infty$. If $0<z<\infty$, as $\Theta \left( 0\right)<\delta z/n<\lim _{q\rightarrow \infty }\Theta \left( q\right)$ and $\Theta \left( q\right)$ is continuous, there exists $0<f\left( z\right) <\infty$ such that $\Theta \left( f\left( z\right) \right) =\delta z/n$ (by the intermediate value theorem). As $\Theta \left( q\right)$ is increasing (since $\Theta '\left( q\right) =\sigma \left( q\right) >0$ for all $q>0$), this solution is unique. $\square$

Lemma 2

For all t, $q\left( t\right) \ge 0$ and $x\left( t\right) \ge 0$.

Proof

By definition, the individual consumption path $c\left( \cdot \right)$ generates trajectories $x\left( \cdot \right)$ and $y\left( \cdot \right)$, such that $\dot{x}\left( t\right) =-nc\left( t\right)$, $x\left( 0\right) =x_{0}$, and $\dot{y}\left( t\right) =\beta n\left( c\left( t\right) -y\left( t\right) \right)$, $y\left( 0\right) =y_{0}$. Thus, letting

$$\begin{aligned} z\left( t\right) =x\left( t\right) -\frac{\alpha }{\left( 1-\alpha \right) \beta }y\left( t\right) , \end{aligned}$$

we can show that

$$\begin{aligned} \dot{z}\left( t\right) =-n\left( \frac{c\left( t\right) -\alpha y\left( t\right) }{1-\alpha }\right) =-nf\left( z\left( t\right) \right) ,\quad z\left( 0\right) =x_{0}-\frac{\alpha }{\left( 1-\alpha \right) \beta }y_{0}\text{. } \end{aligned}$$

Under Condition 1, Lemma 1 implies that $z\left( t\right) \ge 0$, for all t, and $\lim _{t\rightarrow \infty }z\left( t\right) =0$. Now, substituting $c\left( t\right) =\left( 1-\alpha \right) f\left( z\left( t\right) \right) +\alpha y\left( t\right)$, we can obtain

$$\begin{aligned} \dot{y}\left( t\right) =\left( 1-\alpha \right) \beta n\left( f\left( z\left( t\right) \right) -y\left( t\right) \right) ,\quad y\left( 0\right) =y_{0}, \end{aligned}$$

implying that

$$\begin{aligned} y\left( t\right) =\mathrm{e}^{-\left( 1-\alpha \right) \beta nt}\left( \left( 1-\alpha \right) \beta n\int _{0}^{t}\mathrm{e}^{\left( 1-\alpha \right) \beta ns}f\left( z\left( s\right) \right) \mathrm{d}s+y_{0}\right) \text{. } \end{aligned}$$

Clearly, as $y_{0}\ge 0$ and $f\left( z\left( t\right) \right) \ge 0$, for all t, $y\left( t\right) \ge 0$, for all t. Finally, as $y\left( t\right) \ge 0$ and $z\left( t\right) \ge 0$, for all t, we can show that $c\left( t\right) =\left( 1-\alpha \right) f\left( z\left( t\right) \right) +\alpha y\left( t\right) \ge 0$ and $x\left( t\right) =\frac{\alpha }{\left( 1\,-\,\alpha \right) \beta }y\left( t\right) +z\left( t\right) \ge 0$, for all t. $\square$

Remark

Given that $\lim _{t\rightarrow \infty }f\left( z\left( t\right) \right) =0$, we can show that $\lim _{t\rightarrow \infty }y\left( t\right) =0$. It follows that $\lim _{t\rightarrow \infty }x\left( t\right) =\lim _{t\rightarrow \infty }y\left( t\right) =\lim _{t\rightarrow \infty }z\left( t\right) =0$.

(2) Necessary Conditions Let T be the time when $z\left( t\right) =0$ (including the possibility that $T=\infty$). We first construct $\lambda \left( \cdot \right)$, $\mu \left( \cdot \right)$ and $\eta \left( \cdot \right)$ to simultaneously satisfy the maximum condition (5) and the adjoint Eqs. (7) and (8). We then show that the transversality condition (9) holds.

The following lemma will be useful below.

Lemma 3

The marginal utility $u^{\prime }\left( f\left( z\left( t\right) \right) \right)$ grows at the rate $\delta$, for all $t<T$, and is equal to $u'\left( 0\right)$, for all $t\ge T$.

Proof

First remark that $\Theta \left( f\left( z\right) \right) =\delta z/n$, for all z, implies that $\sigma \left( f\left( z\right) \right) f^{\prime }\left( z\right) =\delta /n$, for all z (by differentiation). Define $p\left( t\right) =u^{\prime }\left( f\left( z\left( t\right) \right) \right)$, for all t. If $t<T$, differentiation yields

$$\begin{aligned} \dot{p}\left( t\right) =u''\left( f\left( z\left( t\right) \right) \right) f^{\prime }\left( z\left( t\right) \right) \dot{z}\left( t\right) \text {.} \end{aligned}$$

Substituting $\dot{z}\left( t\right) =-nf\left( z\left( t\right) \right)$ and dividing by $p\left( t\right) =u^{\prime }\left( f\left( z\left( t\right) \right) \right)$, we get

$$\begin{aligned} \frac{\dot{p}\left( t\right) }{p\left( t\right) }=n\sigma \left( f\left( z\left( t\right) \right) \right) f^{\prime }\left( z\left( t\right) \right) =\delta \text {.} \end{aligned}$$

If $t\ge T$, $z\left( t\right) =0$ (by definition) and $f\left( 0\right) =0$ imply that $p\left( t\right) =u^{\prime }\left( 0\right)$. $\square$

For all $t<T$, let

$$\begin{aligned} \lambda \left( t\right) =u^{\prime }\left( f\left( z\left( t\right) \right) \right) ,\quad \mu \left( t\right) =-\frac{\alpha }{\left( 1-\alpha \right) \beta }u^{\prime }\left( f\left( z\left( t\right) \right) \right) \text{ and } \eta \left( t\right) =0\text{. } \end{aligned}$$

(10)

As $u^{\prime }\left( f\left( z\left( t\right) \right) \right) /\left( 1-\alpha \right) =\lambda \left( t\right) -\beta \mu \left( t\right) -\eta \left( t\right) /\left( 1-\alpha \right)$ (by construction), the maximum condition (5) is satisfied if $q\left( t\right) =f\left( z\left( t\right) \right)$. The adjoint Eqs. (7) and (8) are satisfied from Lemma 3.²⁰

For all $t\ge T$, let

$$\begin{aligned} \lambda \left( t\right)= & {} \mathrm{e}^{\delta \left( t-T\right) }u^{\prime }\left( 0\right) ,\nonumber \\ \mu \left( t\right)= & {} -\frac{\alpha }{\left( 1-\alpha \right) \beta }\mathrm{e}^{\delta \left( t-T\right) }u^{\prime }\left( 0\right) \text{ and } \eta \left( t\right) =\left( \mathrm{e}^{\delta \left( t-T\right) }-1\right) u^{\prime }\left( 0\right) \text{. } \end{aligned}$$

(11)

As $u^{\prime }\left( 0\right) /\left( 1-\alpha \right) =\lambda \left( t\right) -\beta \mu \left( t\right) -\eta \left( t\right) /\left( 1-\alpha \right)$ (by construction), the maximum condition (5) is satisfied if $q\left( t\right) =f\left( 0\right) =0$. The adjoint Eqs. (7) and (8) are satisfied by construction.²¹

To verify the transversality condition (9), define $A\left( t\right) =\mathrm{e}^{-\delta t}\left( \lambda \left( t\right) x\left( t\right) +\mu \left( t\right) y\left( t\right) \right)$. We need to separate the cases where T is infinite or finite. If T is infinite, we can show that $A\left( t\right) =u^{\prime }\left( f\left( z_{0}\right) \right) z\left( t\right)$, for all t. Then, the transversality condition follows from $\lim _{t\rightarrow \infty }z\left( t\right) =0$ (see the proof of Lemma 2). If T is finite, we can show that $A\left( t\right) =\mathrm{e}^{-\delta T}u^{\prime }\left( 0\right) z\left( t\right)$, for all $t\ge T$. As $z\left( t\right) =0$ for all $t\ge T$ (by definition of T), the transversality condition is trivially satisfied.

(3) Sufficiency Conditions We show that the Hamiltonian, calculated with the adjoint variables and multiplier defined above, is strictly concave.

For $t<T$, substituting (10), the Hamiltonian can be written after simplification

$$\begin{aligned} H\left( x,y,\lambda ,\mu ,\eta ,c\right) =n\left[ u\left( \frac{c-\alpha y}{1-\alpha }\right) -u^{\prime }\left( f\left( z\left( t\right) \right) \right) \frac{c-\alpha y}{1-\alpha }\right] . \end{aligned}$$

Likewise, for $t\ge T$, substituting (11), we obtain

$$\begin{aligned} H\left( x,y,\lambda ,\mu ,\eta ,c\right) =n\left[ u\left( \frac{c-\alpha y}{1-\alpha }\right) -u^{\prime }\left( 0\right) \frac{c-\alpha y}{1-\alpha }\right] . \end{aligned}$$

Hence, under Assumption 1, the Hamiltonian is strictly concave in $c_{i}$, x and y. This proves that Proposition 1 characterizes the unique optimal solution. $\square$

1.3 Proof of Corollary 1.

Along the cooperative path, we have $q_{i}\left( t\right) =f\left( z\left( t\right) \right)$, where $f\left( z\right)$ is implicitly defined by $\Theta \left( f\left( z\right) \right) =\intop _{0}^{f\left( z\right) }\sigma \left( s\right) \mathrm{d}s=\delta z/n$ and $z\left( t\right)$ satisfies $\dot{z}\left( t\right) =-nf\left( z\left( t\right) \right)$ and $z\left( 0\right) =z_{0}$. Given that $f\left( z\right) >0$ for all $z>0$ (see Lemma 1), it will be sufficient to show that $z\left( t\right) >0$ for all t.

Assume that $\sigma \left( q\right) \ge \mu$ for all q, for some $\mu >0$. It follows that $\Theta \left( q\right) \ge \mu q$. In particular, $\Theta \left( f\left( z\right) \right) \ge \mu f\left( z\right)$. Using $\Theta \left( f\left( z\right) \right) =\delta z/n$, this implies that $f\left( z\right) \le \frac{\delta }{\mu }\frac{z}{n}$ and $\dot{z}\left( t\right) =-nf\left( z\left( t\right) \right) \ge -\frac{\delta }{\mu }z\left( t\right)$. In turn, this allows to show that $z\left( t\right) \ge z_{0}\mathrm{e}^{-\frac{\delta }{\mu }t}$. (Indeed, let $F\left( t\right) =\mathrm{e}^{\frac{\delta }{\mu }t}z\left( t\right)$. Then $F'\left( t\right) =\mathrm{e}^{\frac{\delta }{\mu }t}\left( \dot{z}\left( t\right) +\frac{\delta }{\mu }z\left( t\right) \right) \ge 0$ and $F\left( t\right)$ is increasing. Hence, $F\left( t\right) =\mathrm{e}^{\frac{\delta }{\mu }t}z\left( t\right) \ge z_{0}=F\left( 0\right)$ for all t). Assuming that $z_{0}>0$, this proves that $z\left( t\right) >0$ for all t.

1.4 Proof of Property 1

Denote by $q^{\circ }$ the optimal individual subjective consumption. It satisfies $\Theta \left( q^{\circ }\right) =\frac{\delta }{n}z$, where $z=x-\frac{\alpha }{1-\alpha }\frac{y}{\beta }$. By differentiation, we can obtain

$$\begin{aligned} \frac{\hbox {d}q^{\circ }}{\hbox {d}\alpha }= & {} -\frac{\delta y}{\left( 1-\alpha \right) ^{2}\beta n\sigma \left( q^{\circ }\right) }<0,\\ \frac{\hbox {d}q^{\circ }}{\hbox {d}\beta }= & {} \frac{\alpha \delta y}{\left( 1-\alpha \right) \beta ^{2}n\sigma \left( q^{\circ }\right) }>0,\\ \frac{\hbox {d}q^{\circ }}{\hbox {d}\delta }= & {} \frac{z}{n\sigma \left( q^{\circ }\right) }>0,\\ \frac{\hbox {d}q^{\circ }}{\hbox {d}n}= & {} -\frac{\delta z}{n^{2}\sigma \left( q^{\circ }\right) }>0,\\ \frac{\hbox {d}q^{\circ }}{\hbox {d}x}= & {} \frac{\delta }{n\sigma \left( q^{\circ }\right) }>0,\\ \frac{\hbox {d}q^{\circ }}{\hbox {d}y}= & {} -\frac{\alpha \delta }{\left( 1-\alpha \right) \beta n\sigma \left( q^{\circ }\right) }<0\text{. } \end{aligned}$$

Now, denote by $c^{\circ }$ the optimal individual consumption. It is given by $c^{\circ }=\left( 1-\alpha \right) q^{\circ }+\alpha y$. Using the previous results, we get after differentiation and substitution

$$\begin{aligned} \frac{\hbox {d}c^{\circ }}{\hbox {d}\alpha }= & {} -q^{\circ }-\left( \frac{\delta }{\left( 1-\alpha \right) \beta n\sigma \left( q^{\circ }\right) }-1\right) y,\\ \frac{\hbox {d}c^{\circ }}{\hbox {d}\beta }= & {} \frac{\alpha \delta y}{\beta ^{2}n\sigma \left( q^{\circ }\right) }>0,\\ \frac{\hbox {d}c^{\circ }}{\hbox {d}\delta }= & {} \frac{\left( 1-\alpha \right) z}{n\sigma \left( q^{\circ }\right) }>0,\\ \frac{\hbox {d}c^{\circ }}{\hbox {d}n}= & {} -\frac{\left( 1-\alpha \right) \delta z}{n^{2}\sigma \left( q^{\circ }\right) }<0,\\ \frac{\hbox {d}c^{\circ }}{\hbox {d}x}= & {} \frac{\left( 1-\alpha \right) \delta }{n\sigma \left( q^{\circ }\right) }>0,\\ \frac{\hbox {d}c^{\circ }}{\hbox {d}y}= & {} -\left( \frac{\delta }{\beta n\sigma \left( q^{\circ }\right) }-1\right) \alpha \text{. } \end{aligned}$$

The comparative statics with respect to $\alpha$ is ambiguous. If $\left( 1-\alpha \right) \beta n\sigma \left( q^{\circ }\right) \le \delta$, knowing that $q^{\circ }>0$, we show that $\hbox {d}c^{\circ }/\hbox {d}\alpha <0$. Otherwise, the derivative $\hbox {d}c^{\circ }/\hbox {d}\alpha$ can be negative, zero or positive, depending on $q^{\circ }$ and y. However, let S be the set of all pairs x and y such that $x-\frac{\alpha }{\left( 1\,-\,\alpha \right) \beta }y=z$, as shown in Fig. 7.

Fig. 7

$S=\left\{ (x,y)\in {\mathbb {R}}_{+}^{2};x-\frac{\alpha }{(1-\alpha )\beta }y=z\right\}$

Remark that $q^{\circ }=f\left( z\right)$, for all $\left( x,y\right) \in S$. Thus, since $\left( 1-\alpha \right) \beta n\sigma \left( q^{\circ }\right) >\delta$ and $q^{\circ }$ is constant, $\hbox {d}c^{\circ }/\hbox {d}\alpha$ is linearly decreasing in y, for all $\left( x,y\right) \in S$. This proves that for all $\left( x,y\right) \in S$, there exists $\theta >0$ such that $dc^{\circ }/d\alpha \gtreqless 0$ if and only if $y/x\gtreqless \theta$. This completes the proof of Property 1. $\square$

1.5 Proof of Proposition 2

Consider any player i. Assume that all individuals $j\ne i$ play the stationary Markovian strategy $s_{j}^{*}\left( x,y\right) =\left( 1-\alpha \right) g\left( x-\frac{\alpha }{\left( 1-\alpha \right) \beta }y\right) +\alpha y$ , where $g\left( z\right)$ satisfies

$$\begin{aligned} \Psi \left( g\left( z\right) \right) \equiv \int _{0}^{g\left( z\right) }\left( \sigma \left( s\right) -\left( n-1\right) /n\right) \mathrm{d}s=\frac{\delta }{n}z,\quad \text{ for } \text{ all }\,z\ge 0\text{. } \end{aligned}$$

The problem of the remaining player i is to find a consumption path, $c_{i}\left( \cdot \right)$, maximizing

$$\begin{aligned} \begin{array}{l} \int _{0}^{\infty }u\left( q_{i}\left( t\right) \right) \mathrm{e}^{-\delta t}\mathrm{d}t\text {,}\\ \text {where:}\\ \dot{x}\left( t\right) =-\left( c_{i}\left( t\right) +\sum \limits _{j\ne i}s_{j}^{*}\left( x\left( t\right) ,y\left( t\right) \right) \right) ,\quad x\left( 0\right) =x_{0}\text {,}\\ \dot{y}\left( t\right) =\beta \left( c_{i}\left( t\right) +\sum \limits _{j\ne i}s_{j}^{*}\left( x\left( t\right) ,y\left( t\right) \right) -ny\left( t\right) \right) ,\quad y\left( 0\right) =y_{0}\text {,}\\ q_{i}\left( t\right) \ge 0\quad \text {and}\,x\left( t\right) \ge 0, \end{array} \end{aligned}$$

where we write $q_{i}\left( t\right) =\left( c_{i}\left( t\right) -\alpha y\left( t\right) \right) /\left( 1-\alpha \right)$ to simplify the notations.

Define the current-value Hamiltonian

$$\begin{aligned} H_{i}\left( x,y,\lambda _{i},\mu _{i},\eta _{i},c_{i}\right)&=u\left( q_{i}\right) -\lambda _{i}\left( c_{i}+{\textstyle \sum \limits _{j\ne i}}s_{j}^{*}\left( x,y\right) \right) \\&\quad +\beta \mu _{i}\left( c_{i}+{\textstyle \sum \limits _{j\ne i}}s_{j}^{*}\left( x,y\right) -ny\right) +\eta _{i}q_{i}\\&=u\left( q_{i}\right) -\left( \lambda _{i}-\beta \mu _{i}\right) \left( c_{i}+{\textstyle \sum \limits _{j\ne i}}s_{j}^{*}\left( x,y\right) \right) -\beta n\mu _{i}y+\eta _{i}q_{i} \end{aligned}$$

where $\lambda _{i}$ and $\mu _{i}$ are costate variables associated with the states x and y, and $\eta _{i}$ is a Lagrangian multiplier associated with the feasibility constraint.

If a feasible control path $c_{i}\left( \cdot \right)$, with the corresponding states trajectories $x\left( \cdot \right)$ and $y\left( \cdot \right)$, is optimal, there exists (continuous with piecewise-continuous derivatives) costate variables $\lambda _{i}\left( \cdot \right)$ and $\mu _{i}\left( \cdot \right)$, and a (continuous) multiplier $\eta _{i}\left( \cdot \right)$ such that²²

$$\begin{aligned} \frac{u^{\prime }\left( q_{i}\left( t\right) \right) }{1-\alpha }= & {} \lambda _{i}\left( t\right) -\beta \mu _{i}\left( t\right) -\frac{\eta _{i}\left( t\right) }{1-\alpha }, \end{aligned}$$

(12)

$$\begin{aligned} \eta _{i}\left( t\right)\ge & {} 0,\quad q_{i}\left( t\right) \ge 0\quad \text{ and }\,\eta _{i}\left( t\right) q_{i}\left( t\right) =0, \end{aligned}$$

(13)

$$\begin{aligned} \dot{\lambda }_{i}\left( t\right)= & {} \delta \lambda _{i}\left( t\right) +\left( \lambda _{i}\left( t\right) -\beta \mu _{i}\left( t\right) \right) \left( n-1\right) \left( 1-\alpha \right) g'\left( z\left( t\right) \right) , \end{aligned}$$

(14)

$$\begin{aligned} \dot{\mu }_{i}\left( t\right)= & {} \left( \delta +\beta n\right) \mu _{i}\left( t\right) +\left( \lambda _{i}\left( t\right) -\beta \mu _{i}\left( t\right) \right) \left( n-1\right) \frac{\alpha }{\beta }\left( \beta -g'\left( z\left( t\right) \right) \right) \nonumber \\&\quad +\frac{\alpha }{1-\alpha }\left( u^{\prime }\left( q_{i}\left( t\right) \right) +\eta _{i}\left( t\right) \right) \end{aligned}$$

(15)

$$\begin{aligned}&\lim _{t\rightarrow \infty }\mathrm{e}^{-\delta t}\left( \lambda _{i}\left( t\right) x\left( t\right) +\mu _{i}\left( t\right) y\left( t\right) \right) =0, \end{aligned}$$

(16)

where we write $z\left( t\right) =x\left( t\right) -\frac{\alpha }{\left( 1-\alpha \right) \beta }y\left( t\right)$.

Now, let $c_{i}\left( t\right)$, for all t, satisfy

$$\begin{aligned} c_{i}\left( t\right) =\left( 1-\alpha \right) g\left( x\left( t\right) -\frac{\alpha }{\left( 1-\alpha \right) \beta }y\left( t\right) \right) +\alpha y\left( t\right) , \end{aligned}$$

where $x\left( t\right)$ and $y\left( t\right)$ are the corresponding states trajectories. Remark that it is equivalent to write

$$\begin{aligned} q_{i}\left( t\right) =g\left( x\left( t\right) -\frac{\alpha }{\left( 1-\alpha \right) \beta }y\left( t\right) \right) \text{. } \end{aligned}$$

We show below that the proposed consumption path $c_{i}\left( \cdot \right)$ is feasible, and we find $\lambda _{i}\left( \cdot \right)$, $\mu _{i}\left( \cdot \right)$ and $\eta _{i}\left( \cdot \right)$, to satisfy conditions (12) to (16). Proposition 2 follows.

(1) Feasibility We first show, in Lemmas 4 and 5, that the proposed control path $c_{i}\left( \cdot \right)$ is feasible.

Lemma 4

For all $z>0$, $g\left( z\right) >0$, and $g\left( 0\right) =0$.

Proof

If $z=0$, $g\left( 0\right) =0$ follows from $\Psi \left( 0\right) =0$. Likewise, $\lim _{z\rightarrow \infty }g\left( z\right) =\infty$ follows from $\lim _{q\rightarrow \infty }\Psi \left( q\right) =\infty$. If $0<z<\infty$, as $\Psi \left( 0\right)<\delta z/n<\lim _{q\rightarrow \infty }\Psi \left( q\right)$ and $\Psi \left( q\right)$ is continuous, there exists $0<g\left( z\right) <\infty$ such that $\Psi \left( g\left( z\right) \right) =\delta z/n$ (by the intermediate value theorem). As $\Psi \left( q\right)$ is increasing (assuming $\Psi '\left( q\right) =\sigma \left( q\right) -\left( n-1\right) /n>0$ for all $q>0$), this solution is unique. $\square$

Lemma 5

For all t, $q_{i}\left( t\right) \ge 0$ and $x\left( t\right) \ge 0$.

Proof

By definition, the individual consumption path $c_{i}\left( \cdot \right)$ generates trajectories $x\left( \cdot \right)$ and $y\left( \cdot \right)$, such that $\dot{x}\left( t\right) =-\sum \limits _{j=1}^{n}c_{j}\left( t\right)$, $x\left( 0\right) =x_{0}$, and $\dot{y}\left( t\right) =\beta \left( \sum \limits _{j=1}^{n}c_{j}\left( t\right) -ny\left( t\right) \right)$, $y\left( 0\right) =y_{0}$. Thus, letting

$$\begin{aligned} z\left( t\right) =x\left( t\right) -\frac{\alpha }{\left( 1-\alpha \right) \beta }y\left( t\right) , \end{aligned}$$

we can show that

$$\begin{aligned} \dot{z}\left( t\right) =-\sum \limits _{j=1}^{n}\frac{c_{j}\left( t\right) -\alpha y\left( t\right) }{1-\alpha }=-ng\left( z\left( t\right) \right) ,\quad z\left( 0\right) =x_{0}-\frac{\alpha }{\left( 1-\alpha \right) \beta }y_{0}\text{. } \end{aligned}$$

Under Condition 1, Lemma 4 implies that $z\left( t\right) \ge 0$, for all t, and $\lim _{t\rightarrow \infty }z\left( t\right) =0$. Now, substituting $c_{j}\left( t\right) =\left( 1-\alpha \right) g\left( z\left( t\right) \right) +\alpha y\left( t\right)$, for all j, we can obtain

$$\begin{aligned} \dot{y}\left( t\right) =\left( 1-\alpha \right) \beta n\left( g\left( z\left( t\right) \right) -y\left( t\right) \right) ,\quad y\left( 0\right) =y_{0}, \end{aligned}$$

implying that

$$\begin{aligned} y\left( t\right) =\mathrm{e}^{-\left( 1-\alpha \right) \beta nt}\left( \left( 1-\alpha \right) \beta n\int _{0}^{t}\mathrm{e}^{\left( 1-\alpha \right) \beta ns}g\left( z\left( s\right) \right) \mathrm{d}s+y_{0}\right) \text{. } \end{aligned}$$

Clearly, as $y_{0}\ge 0$ and $g\left( z\left( t\right) \right) \ge 0$, for all t, $y\left( t\right) \ge 0$, for all t. Finally, as $y\left( t\right) \ge 0$ and $z\left( t\right) \ge 0$, for all t, we can show that $c\left( t\right) =\alpha y\left( t\right) +\left( 1-\alpha \right) g\left( z\left( t\right) \right) \ge 0$ and $x\left( t\right) =\frac{\alpha }{\left( 1-\alpha \right) \beta }y\left( t\right) +z\left( t\right) \ge 0$, for all t. $\square$

Remark

Given that $\lim _{t\rightarrow \infty }g\left( z\left( t\right) \right) =0$, $\lim _{t\rightarrow \infty }y\left( t\right) =0$. Hence, we can show that $\lim _{t\rightarrow \infty }x\left( t\right) =\lim _{t\rightarrow \infty }y\left( t\right) =\lim _{t\rightarrow \infty }z\left( t\right) =0$.

(2) Necessary Conditions Let T be the time when $z\left( t\right) =0$ (including the possibility that $T=\infty$). We first construct $\lambda _{i}\left( \cdot \right)$, $\mu _{i}\left( \cdot \right)$ and $\eta _{i}\left( \cdot \right)$ to simultaneously satisfy the maximum condition (12) and the adjoint Eqs. (14) and (15). We then show that the transversality condition (16) holds.

The following lemma will be useful below.

Lemma 6

The marginal utility $u^{\prime }\left( g\left( z\left( t\right) \right) \right)$ grows at the rate $\delta +\left( n-1\right) g'\left( z\left( t\right) \right)$, for all $t<T$, and is equal to $u'\left( 0\right)$, for all $t\ge T$.

Proof

First remark that $\Psi \left( g\left( z\right) \right) =\delta z/n$, for all z, implies that $\sigma \left( g\left( z\right) \right) g'\left( z\right) =\left( \delta +\left( n-1\right) g'\left( z\right) \right) /n$, for all z (by differentiation). Define $p\left( t\right) =u'\left( g\left( z\left( t\right) \right) \right)$, for all t. If $t<T$, differentiation yields

$$\begin{aligned} \dot{p}\left( t\right) =u''\left( g\left( z\left( t\right) \right) \right) g'\left( z\left( t\right) \right) \dot{z}\left( t\right) \text {.} \end{aligned}$$

Substituting $\dot{z}\left( t\right) =-ng\left( z\left( t\right) \right)$ and dividing by $p\left( t\right) =u^{\prime }\left( g\left( z\left( t\right) \right) \right)$, we get

$$\begin{aligned} \frac{\dot{p}\left( t\right) }{p\left( t\right) }=n\sigma \left( g\left( z\left( t\right) \right) \right) g'\left( z\left( t\right) \right) =\delta +\left( n-1\right) g'\left( z\left( t\right) \right) \text {.} \end{aligned}$$

If $t\ge T$, $z\left( t\right) =0$ (by definition) and $g\left( 0\right) =0$ imply that $p\left( t\right) =u^{\prime }\left( 0\right)$. $\square$

For all $t<T$, let

$$\begin{aligned} \lambda _{i}\left( t\right) =u'\left( g\left( z\left( t\right) \right) \right) ,\quad \mu _{i}\left( t\right) =-\frac{\alpha }{\left( 1-\alpha \right) \beta }u'\left( g\left( z\left( t\right) \right) \right) \text{ and } \eta _{i}\left( t\right) =0\text{. } \end{aligned}$$

(17)

As $u'\left( g\left( z\left( t\right) \right) \right) /\left( 1-\alpha \right) =\lambda _{i}\left( t\right) -\beta \mu _{i}\left( t\right) -\eta _{i}\left( t\right) /\left( 1-\alpha \right)$ (by construction), the maximum condition (12) is satisfied if $q_{i}\left( t\right) =g\left( z\left( t\right) \right)$. The adjoint Eqs. (14) and (15) are satisfied from Lemma 6.²³

For all $t\ge T$, let

$$\begin{aligned} \lambda _{i}\left( t\right) =\mathrm{e}^{\rho \left( t-T\right) }u^{\prime }\left( 0\right) ,\quad \mu _{i}\left( t\right) =-\frac{\alpha }{\left( 1-\alpha \right) \beta }\mathrm{e}^{\rho \left( t-T\right) }u^{\prime }\left( 0\right) \text{ and } \eta _{i}\left( t\right) =\left( \mathrm{e}^{\rho \left( t-T\right) }-1\right) u^{\prime }\left( 0\right) ,\nonumber \\ \end{aligned}$$

(18)

where we let $\rho \equiv \delta +\left( n-1\right) g'\left( 0\right)$. As $u^{\prime }\left( 0\right) /\left( 1-\alpha \right) =\lambda _{i}\left( t\right) -\beta \mu _{i}\left( t\right) -\eta _{i}\left( t\right) /\left( 1-\alpha \right)$ (by construction), the maximum condition (12) is satisfied if $q_{i}\left( t\right) =g\left( 0\right) =0$. The adjoint Eqs. (14) and (15) are satisfied by construction.²⁴

To verify the transversality condition (16), define $A_{i}\left( t\right) =\mathrm{e}^{-\delta t}\left( \lambda _{i}\left( t\right) x\left( t\right) +\mu _{i}\left( t\right) y\left( t\right) \right)$. We need to separate the cases where T is infinite or finite. If T is infinite, we can show that $A_{i}\left( t\right) =\mathrm{e}^{-\delta t}u'\left( g\left( z\left( t\right) \right) \right) z\left( t\right)$ for all t. By Lemma 6, we can get by differentiation

$$\begin{aligned} \frac{\dot{A}_{i}\left( t\right) }{A_{i}\left( t\right) }=\left( n-1\right) g'\left( z\left( t\right) \right) -n\frac{g\left( z\left( t\right) \right) }{z\left( t\right) }\text{. } \end{aligned}$$

Since $\lim _{t\rightarrow \infty }z\left( t\right) =0$ (see the proof of Lemma 5) and $g\left( 0\right) =0$, we have

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{g\left( z\left( t\right) \right) }{z\left( t\right) }=\lim _{\varepsilon \rightarrow 0}\frac{g\left( \varepsilon \right) -g\left( 0\right) }{\varepsilon }=g'\left( 0\right) , \end{aligned}$$

which implies that

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{\dot{A}_{i}\left( t\right) }{A_{i}\left( t\right) }=-g'\left( 0\right) \text{. } \end{aligned}$$

As $g'\left( 0\right) >0$ by assumption, it follows that $\lim _{t\rightarrow \infty }A_{i}\left( t\right) =0$. If T is finite, we can show that $A_{i}\left( t\right) =\mathrm{e}^{-\delta t}\mathrm{e}^{\rho \left( t-T\right) }u^{\prime }\left( 0\right) z\left( t\right)$ for all $t\ge T$. As $z\left( t\right) =0$ for all $t\ge T$ (by definition of T), the transversality condition is trivially satisfied.

(3) Sufficiency conditionsWe show that the Hamiltonian, calculated with the adjoint variables and multiplier defined above, is strictly concave.

For $t<T$, using $s_{j}^{*}\left( x,y\right) =\left( 1-\alpha \right) g\left( x-\frac{\alpha }{\left( 1-\alpha \right) \beta }y\right) +\alpha y$ and substituting (17), the Hamiltonian can be written after simplification

$$\begin{aligned}&H_{i}\left( x,y,\lambda _{i},\mu _{i},\eta _{i},c_{i}\right) \\&\quad =u\left( \frac{c_{i}-\alpha y}{1-\alpha }\right) -u'\left( g\left( z\left( t\right) \right) \right) \frac{c_{i}-\alpha y}{1-\alpha }-u'\left( g\left( z\left( t\right) \right) \right) \left( n-1\right) g\left( x-\frac{\alpha }{\left( 1-\alpha \right) \beta }y\right) . \end{aligned}$$

Likewise, for $t\ge T$, using $s_{j}^{*}\left( x,y\right) =\left( 1-\alpha \right) g\left( x-\frac{\alpha }{\left( 1-\alpha \right) \beta }y\right) +\alpha y$ and substituting (18), we obtain

$$\begin{aligned}&H_{i}\left( x,y,\lambda _{i},\mu _{i},\eta _{i},c_{i}\right) \\&\quad =u\left( \frac{c_{i}-\alpha y}{1-\alpha }\right) -u^{\prime }\left( 0\right) \frac{c_{i}-\alpha y}{1-\alpha }-\mathrm{e}^{\rho \left( t-T\right) }u^{\prime }\left( 0\right) \left( n-1\right) g\left( x-\frac{\alpha }{\left( 1-\alpha \right) \beta }y\right) . \end{aligned}$$

Using $\Psi \left( g\left( z\right) \right) =\delta z/n$, for all z, we can show by differentiation that $g'\left( z\right) =\delta /\left( n\sigma \left( g\left( z\right) \right) -n+1\right)$ and $g''\left( z\right) =-\delta ^{2}/\left( \sigma \left( g\left( z\right) \right) -\left( n-1\right) /n\right) ^{3}<0$, for all z. Hence, under assumption 1, the Hamiltonian is strictly concave in $c_{i}$, x and y. This proves that the strategy considered above is the unique best-reply of player i to the strategies $s_{j}^{*}\left( x,y\right)$ played by the other players. $\square$

1.6 Proof of Corollary 2.

Along the noncooperative path, we have $q_{i}\left( t\right) =g\left( z\left( t\right) \right)$, where $g\left( z\right)$ is implicitly defined by $\Psi \left( g\left( z\right) \right) \equiv \int _{0}^{g\left( z\right) }\left( \sigma \left( s\right) -\left( n-1\right) /n\right) \mathrm{d}s=\frac{\delta }{n}z$ and $z\left( t\right)$ satisfies $\dot{z}\left( t\right) =-ng\left( z\left( t\right) \right)$ and $z\left( 0\right) =z_{0}$. Given that $g\left( z\right) >0$ for all $z>0$ (see Lemma 4), it will be sufficient to show that $z\left( t\right) >0$ for all t.

Assume that $\sigma \left( q\right) -\left( n-1\right) /n\ge \mu$ for all q, for some $\mu >0$. It follows that $\Psi \left( q\right) \ge \mu q$. In particular, $\Psi \left( f\left( z\right) \right) \ge \mu f\left( z\right)$. Using $\Psi \left( f\left( z\right) \right) =\delta z/n$, this implies that $g\left( z\right) \le \frac{\delta }{\mu }\frac{z}{n}$ and $\dot{z}\left( t\right) =-ng\left( z\left( t\right) \right) \ge -\frac{\delta }{\mu }z\left( t\right)$. In turn, this allows to show that $z\left( t\right) \ge z_{0}\mathrm{e}^{-\frac{\delta }{\mu }t}$.(Indeed, let $F\left( t\right) =\mathrm{e}^{\frac{\delta }{\mu }t}z\left( t\right)$. Then $F'\left( t\right) =\mathrm{e}^{\frac{\delta }{\mu }t}\left( \dot{z}\left( t\right) +\frac{\delta }{\mu }z\left( t\right) \right) \ge 0$ and $F\left( t\right)$ is increasing. Hence, $F\left( t\right) =\mathrm{e}^{\frac{\delta }{\mu }t}z\left( t\right) \ge z_{0}=F\left( 0\right)$ for all t.) Assuming that $z_{0}>0$, this proves that $z\left( t\right) >0$ for all t.

1.7 Proof of Property 2

Denote by $q^{*}$ the Nash equilibrium individual subjective consumption. It satisfies $\Psi \left( q^{*}\right) =\Theta \left( q^{*}\right) -\frac{n-1}{n}q^{*}=\frac{\delta }{n}z$, where $z=x-\frac{\alpha }{\left( 1\,-\,\alpha \right) \beta }y$. Remember that the elasticity of the marginal utility $\sigma \left( \cdot \right)$ is assumed larger than $\left( n-1\right) /n$. By differentiation, we can obtain

$$\begin{aligned} \frac{\mathrm{d}q^{*}}{\mathrm{d}\alpha }= & {} -\frac{\delta y}{\left( 1-\alpha \right) ^{2}\beta n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }<0, \\ \frac{\mathrm{d}q^{*}}{\mathrm{d}\beta }= & {} \frac{\alpha \delta y}{\left( 1-\alpha \right) \beta ^{2}n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }>0, \\ \frac{\mathrm{d}q^{*}}{\mathrm{d}\delta }= & {} \frac{z}{n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }>0, \\ \frac{\mathrm{d}q^{*}}{\mathrm{d}n}= & {} -\frac{\Theta \left( q^{*}\right) -q^{*}}{n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }, \\ \frac{\mathrm{d}q^{*}}{\mathrm{d}x}= & {} \frac{\delta }{n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }>0, \\ \frac{\mathrm{d}q^{*}}{\mathrm{d}y}= & {} -\frac{\alpha \delta }{\left( 1-\alpha \right) \beta n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }<0\text{. } \end{aligned}$$

The comparative statics with respect to n is ambiguous. However, by the mean value theorem, given that $\Theta \left( 0\right) =0$, there exists q, with $0<q<q^{*}$, such that

$$\begin{aligned} \Theta \left( q^{*}\right) =\sigma \left( q\right) q^{*}\text{. } \end{aligned}$$

Assume that $\sigma \left( q\right) >1$ for all q. Then, $\Theta \left( q^{*}\right) >q^{*}$, implying that $\mathrm{d}q^{*}/\mathrm{d}n<0$. Likewise, one can prove that $\mathrm{d}q^{*}/\mathrm{d}n=0$ if $\sigma \left( q\right) =1$, for all q, and $\mathrm{d}q^{*}/\mathrm{d}n>0$ if $\sigma \left( q\right) <1$, for all q.

Now, denote by $c^{*}$ the optimal individual consumption. It is given by $c^{*}=\left( 1-\alpha \right) q^{*}+\alpha y$. Using the previous results, we get after differentiation and substitution

$$\begin{aligned} \frac{\mathrm{d}c^{*}}{\mathrm{d}\alpha }= & {} -q^{*}-\left( \frac{\delta }{\left( 1-\alpha \right) \beta n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }-1\right) y_{0},\\ \frac{\hbox {d}c^{*}}{\mathrm{d}\beta }= & {} \frac{\alpha \delta y}{\beta ^{2}n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }>0,\\ \frac{\hbox {d}c^{*}}{\mathrm{d}\delta }= & {} \frac{\left( 1-\alpha \right) z}{n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }>0,\\ \frac{\hbox {d}c^{*}}{\mathrm{d}n}= & {} -\frac{\left( 1-\alpha \right) \left( \Theta \left( q^{*}\right) -q^{*}\right) }{n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) },\\ \frac{\hbox {d}c^{*}}{\mathrm{d}x}= & {} \frac{\left( 1-\alpha \right) \delta }{n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }>0,\\ \frac{\hbox {d}c^{*}}{\mathrm{d}y}= & {} -\left( \frac{\delta }{\beta n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) }-1\right) \alpha \text{. } \end{aligned}$$

The comparative statics with respect to $\alpha$ is ambiguous. If $\left( 1-\alpha \right) \beta n\left( \sigma \left( q^{*}\right) -\frac{n-1}{n}\right) \le \delta$, knowing that $q^{*}>0$, we show that $\mathrm{d}c^{*}/\mathrm{d}\alpha <0$. Otherwise, the derivative $\mathrm{d}c^{*}/\mathrm{d}\alpha$ can be negative, zero or positive, depending on $q^{*}$ and $y_{0}$. Indeed, let S be the set of all pairs x and y such that $x-\frac{\alpha }{1-\alpha }\frac{y}{\beta }=z$, as drawn in Fig. 7. Remark that $q^{*}=g\left( z\right)$, for all $\left( x,y\right) \in S$. Thus, since $\left( 1-\alpha \right) \beta n\left( \sigma \left( q^{*}\right) -\left( n-1\right) /n\right) >\delta$ and $q^{*}$ is constant, $dc^{*}/d\alpha$ is linearly decreasing in y. This proves that for all $\left( x,y\right) \in S$, there exists $\theta >0$ such that $dc^{*}/d\alpha \gtreqless 0$ if and only if $y/x\gtreqless \theta$. Using the same argument as for $q^{*}$, we can show that $dc^{*}/dn\gtreqless 0$ when $\sigma \left( q^{*}\right) \lesseqgtr 1$. This completes the proof of Property 2. $\square$

1.8 Proof of Property 3

Using results from the proofs of Properties 1 and 2, we can obtain

$$\begin{aligned} \frac{\mathrm{d}\Delta }{\mathrm{d}\alpha }= & {} -\left( q^{*}-q^{\circ }\right) -\left( \frac{1}{\sigma \left( q^{*}\right) -\frac{n-1}{n}}-\frac{1}{\sigma \left( q^{\circ }\right) }\right) \frac{\delta y}{\left( 1-\alpha \right) \beta n},\\ \frac{\mathrm{d}\Delta }{\mathrm{d}\beta }= & {} \left( \frac{1}{\sigma \left( q^{*}\right) -\frac{n-1}{n}}-\frac{1}{\sigma \left( q^{\circ }\right) }\right) \frac{\alpha \delta y}{\beta ^{2}n},\\ \frac{\mathrm{d}\Delta }{\mathrm{d}y_{0}}= & {} -\left( \frac{1}{\sigma \left( q^{*}\right) -\frac{n-1}{n}}-\frac{1}{\sigma \left( q^{\circ }\right) }\right) \frac{\alpha \delta }{\beta n}\text{. } \end{aligned}$$

It follows that $\sigma \left( q^{*}\right) -\sigma \left( q^{\circ }\right) \le \left( n-1\right) /n$ implies $\mathrm{d}\Delta /\mathrm{d}\alpha <0$ (using $q^{*}>q^{\circ }$). It is immediate that $\mathrm{d}\Delta /\mathrm{d}\beta \gtreqless 0$ and $\mathrm{d}\Delta /\mathrm{d}y\lesseqgtr 0$ if and only if $\sigma \left( q^{*}\right) -\sigma \left( q^{\circ }\right) \lesseqgtr \left( n-1\right) /n$.

Given that $q^{*}>q^{\circ }$, it is easy to see that $\sigma \left( q^{*}\right) -\sigma \left( q^{\circ }\right) <\left( n-1\right) /n$ will be true if the elasticity of marginal utility is nonincreasing. It will also be true if $\left| \sigma \left( q\right) -\sigma \left( q'\right) \right| <\left( n-1\right) /n$, for all q and $q'$. $\square$